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Date May Specimen paper Marks available 2 Reference code SPM.2.AHL.TZ0.11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number 11 Adapted from N/A

Question

A large tank initially contains pure water. Water containing salt begins to flow into the tank The solution is kept uniform by stirring and leaves the tank through an outlet at its base. Let x grams represent the amount of salt in the tank and let t minutes represent the time since the salt water began flowing into the tank.

The rate of change of the amount of salt in the tank, dxdt, is described by the differential equation dxdt=10et4xt+1.

Show that t + 1 is an integrating factor for this differential equation.

[2]
a.

Hence, by solving this differential equation, show that x(t)=20040et4(t+5)t+1.

[8]
b.

Sketch the graph of x versus t for 0 ≤ t ≤ 60 and hence find the maximum amount of salt in the tank and the value of t at which this occurs.

[5]
c.

Find the value of t at which the amount of salt in the tank is decreasing most rapidly.

[2]
d.

The rate of change of the amount of salt leaving the tank is equal to xt+1.

Find the amount of salt that left the tank during the first 60 minutes.

[4]
e.

Markscheme

METHOD 1

I(t)=eP(t)dt      M1

e1t+1dt

= eln(t+1)       A1

=t+1       AG

 

METHOD 2

attempting product rule differentiation on ddt(x(t+1))      M1

ddt(x(t+1))=dxdt(t+1)+x

=(t+1)(dxdt+xt+1)       A1

so t+1 is an integrating factor for this differential equation        AG

 

[2 marks]

a.

 

attempting to multiply through by (t+1) and rearrange to give      (M1)

(t+1)dxdt+x=10(t+1)et4         A1

ddt(x(t+1))=10(t+1)et4

x(t+1)=10(t+1)et4dt        A1

attempting to integrate the RHS by parts         M1

=40(t+1)et4+40et4dt

=40(t+1)et4160et4+C         A1

Note: Condone the absence of C.

 

EITHER

substituting t=0,x=0C=200            M1

x=40(t+1)et4160et4+200t+1        A1

using 40et4 as the highest common factor of 40(t+1)et4 and 160et4            M1

 

OR

using 40et4 as the highest common factor of 40(t+1)et4 and 160et4 giving

x(t+1)=40et4(t+5)+C (or equivalent)              M1A1

substituting t=0,x=0C=200            M1

 

THEN

x(t)=20040et4(t+5)t+1        AG

 

[8 marks]

b.

 

graph starts at the origin and has a local maximum (coordinates not required)      A1

sketched for 0 ≤ t ≤ 60      A1

correct concavity for 0 ≤ t ≤ 60      A1

maximum amount of salt is 14.6 (grams) at t = 6.60 (minutes)       A1A1 

[5 marks]

c.

using an appropriate graph or equation (first or second derivative)      M1

amount of salt is decreasing most rapidly at t = 12.9 (minutes)      A1

[2 marks]

d.

EITHER

attempting to form an integral representing the amount of salt that left the tank     M1

600x(t)t+1dt

60020040et4(t+5)(t+1)2dt    A1

 

OR

attempting to form an integral representing the amount of salt that entered the tank minus the amount of salt in the tank at t = 60(minutes)

amount of salt that left the tank is 60010et4dtx(60)    A1

 

THEN

= 36.7 (grams)    A2

[4 marks]

e.

Examiners report

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Syllabus sections

Topic 5 —Calculus » SL 5.5—Integration introduction, areas between curve and x axis
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Topic 5 —Calculus » AHL 5.18—1st order DE’s – Euler method, variables separable, integrating factor, homogeneous DE using sub y=vx
Topic 5 —Calculus

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